$\dfrac{dy}{dt}=y+1$ and $y(0)=3$. What is $t$ when $y=1$ ? Choose 1 answer: Choose 1 answer: (Choice A) A $t=\ln 4$ (Choice B) B $t=\ln 2$ (Choice C) C $t=1$ (Choice D) D $t=0$ (Choice E) E $t=-\ln 2$
Answer: The differential equation is separable. What does it look like after we separate the variables? $\dfrac{dy}{y+1}=dt$ Let's integrate both sides of the equation. $\int\dfrac{dy}{y+1}=\int dt$ What do we get? $\ln|y+1|=t + C$ What value of $C$ satisfies the initial condition $y(0)=3$ ? Let's substitute $t=0$ and $y=3$ into the equation and solve for $C$. $\begin{aligned} \ln|3+1|&=0 + C\\ \\ C&=\ln 4 \end{aligned}$ Now use this value of $C$ to find $t$ when $y=1$. $\begin{aligned} \ln|1+1|&=t + \ln 4\\ \\ t&=\ln 2-\ln 4\\ \\ t&=\ln 2-2\ln 2\\ \\ t&=-\ln 2 \end{aligned}$